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I'm going in ellipses?

  1. Dec 4, 2009 #1
    I'm trying to find the equation of motion of a bead, which is constrained to roll along the bottom half of this frictionless ellipse, by using langrange's method - [tex]L(\phi,\dot{\phi})[/tex].
    Here's the setup:
    given the bottom half of an ellipse:
    [tex]\mathbf{r}(\phi) = acos(\phi)\mathbf{\hat{i}}-bsin(\phi)\mathbf{\hat{j}}[/tex]
    where [tex]0<\phi<\pi[/tex], and [tex]\mathbf{\hat{i}}[/tex] , [tex]\mathbf{\hat{j}}[/tex] are the unit vectors for x and y, and [tex]\phi[/tex] is measured from the positive x axis counterclockwise.

    my question is should I use

    1) [tex]KE=\frac{m}{2}(\dot{r}^{2}+r^{2}\dot{\phi}^{2})[/tex]
    or
    2) [tex]KE=\frac{m}{2}(\dot{x}^{2}+\dot{y}^{2})[/tex]

    where [tex]\dot{x} \:\&\: \dot{y}[/tex] come from [tex]x=acos(\phi)[/tex] , and [tex] y=-bsin(\phi)[/tex]

    which one?

    I'm confused because I have polar type coordinates, [tex]r(\phi)[/tex] in terms of Cartesian components.


    Assuming that the second equation for KE is not appropriate to use here,

    If I use the first the first equation for KE, then is it 'Legal' to find and use [tex] \mathbf{\dot{r}}[/tex] like this? or is this wrong?

    [tex]\mathbf{\dot{r}}=-a\dot{\phi}sin(\phi)\mathbf{\hat{i}}-b\dot{\phi}cos(\phi)\mathbf{\hat{j}}[/tex]

    --------------------------------------------------------------------------------------------------------------
    As my only other option I'm aware of, I know that the definitely correct, brute force approach looks like this:
    [tex]\mathbf{r}(\phi)_{polar}=|\mathbf{r}(\phi)_{cartesian}|\mathbf{\hat{r}}=\sqrt{a^{2}cos^{2}(\phi)+b^{2}sin^{2}(\phi)}\mathbf{\hat{r}}[/tex]

    which means we're in polar coordinates, so

    [tex]\mathbf{\dot{r}}(\phi)=\dot{r}\mathbf{\hat{r}}+r\dot{\phi}\mathbf{\hat{\phi}}[/tex]

    [tex]\mathbf{\dot{r}}(\phi)=\frac{d(\sqrt{a^{2}cos^{2}(\phi)+b^{2}sin^{2}(\phi)})}{dt}\mathbf{\hat{r}}+\dot{\phi}\sqrt{a^{2}cos^{2}(\phi)+b^{2}sin^{2}(\phi)}\mathbf{\hat{\phi}}[/tex]

    and L=KE-PE... and then I have to solve [tex]\frac{dL}{d\phi}-\frac{d(\frac{dL}{d\dot{\phi}})}{dt}=0[/tex]
    which, that route is becoming a nightmare very quickly, and I don't want any trouble with the algebra police... any advice?
     
  2. jcsd
  3. Dec 5, 2009 #2

    Dale

    Staff: Mentor

    The idea in these types of problems is to find an expression for the KE purely in terms of your generalized coordinates. Sometimes you may know the KE immediately in those terms, but other times you may have to start from Cartesian coordinates where you know the expression for KE and then you can transform to your generalized coordinates.

    That is a long-winded way of saying that I would start with 2) and substitute in the a cos and b sin terms.
     
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